Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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Assuming that the problem is well-posed, it should be true in the particular case where <math>S_1 \cong S_3</math> and <math>R_1 \cong R_2</math>. Let the sum of the side lengths of <math>S_1</math> and <math>S_2</math> be <math>x</math>, and let the length of rectangle <math>R_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}</cmath> which we solve to determine <math>y=\boxed{\textbf{(A) }651}</math>. | Assuming that the problem is well-posed, it should be true in the particular case where <math>S_1 \cong S_3</math> and <math>R_1 \cong R_2</math>. Let the sum of the side lengths of <math>S_1</math> and <math>S_2</math> be <math>x</math>, and let the length of rectangle <math>R_2</math> be <math>y</math>. We then have the system <cmath>\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}</cmath> which we solve to determine <math>y=\boxed{\textbf{(A) }651}</math>. | ||
| − | ==Solution 3 ( | + | ==Solution 3 (faster version of solution 1)== |
Since, for each pair of rectangles, the side lengths have a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, the answer must be <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>. | Since, for each pair of rectangles, the side lengths have a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, the answer must be <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{\textbf{(A) }651}</math>. | ||
Revision as of 21:23, 25 November 2020
Contents
Problem
Rectangles 
and 
and squares 
and 
shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of 
in units?
![[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]](http://latex.artofproblemsolving.com/a/5/5/a55c518a1179d67520c6a03827b4ba25ca1413f2.png)
Solution 1
Let the side length of each square 
be 
. Then, from the diagram, we can line up the top horizontal lengths of 
, , and 
to cover the top side of the large rectangle, so 
. Similarly, the short side of 
will be 
, and lining this up with the left side of to cover the vertical side of the large rectangle gives 
. We subtract the second equation from the first to obtain 
, and thus 
.
Solution 2
Assuming that the problem is well-posed, it should be true in the particular case where 
and 
. Let the sum of the side lengths of and be 
, and let the length of rectangle be 
. We then have the system ![\[\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}\]](http://latex.artofproblemsolving.com/b/5/a/b5ad6916616eee9187590637ee867994d51286aa.png)
which we solve to determine 
.
Solution 3 (faster version of solution 1)
Since, for each pair of rectangles, the side lengths have a sum of 
or 
and a difference of , the answer must be 
.
Video Solutions
See also
| 2020 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
